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If $\lim_{x\rightarrow\infty}f(x) = A$, prove that $f(x)\equiv A$.

Any help would be appreciated! I just don't even understand the question.

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    it means that $f(x)$ is $a$ for all $x$, so it is a constant function. Is it possible that $f$ in this case is a polynomial or something? Clearly some context is missing. – 2017-02-08
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    Usually, it means that $f$ is identically equal to $A$. Certainly this question is missing context though, as what is being asked to be proved is not true in general. – 2017-02-08
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    This statement means that $f(x) $ is identically equal to $A$, i.e. $f(x) =A$ for every $x$. – 2017-02-08
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    Can mean "is identically equal to", meaning that the equality holds for all $x$. But this doesn't make sense here (as the proposition would be false). Aren't you missing a part of the problem statement ? – 2017-02-08
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    I think the question probably is about to show that one can define a congruence class using the limiting behavior? – 2017-02-08
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    @YvesDaoust Can you provide a counter example? I am having trouble too see the falseness of the statement – 2017-02-09
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    @ThePortakal: you silently modified the problem statement, didn't you ? – 2017-02-09
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    @YvesDaoust No I did not. I just ran into this question and wondered what could be a counter example. So, you are saying that this is a true statement now? – 2017-02-09
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    @ThePortakal: ooops, sorry, the OP - not you - silently fixed the question (see my first comment). Now the comment is obsolete. – 2017-02-09

2 Answers 2

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For any $a\in (0,\infty)$, the constant $f(a)=f(2^n a)\to A$ as $n\to\infty$. Done.

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    I don't think I understand your answer.. – 2017-02-08
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    @JDel1996, for any $a\in(0,\infty)$, you can define sequence $x_n = a\cdot2^{n-1}$. Note that condition $f(2x) = f(x)$ implies that $f(x_n) = f(a)$ is a constant sequence. On, the other hand, by definition (characterization) of [limit of function in terms of sequences](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences), because $x_n\to\infty$, we have that $A=\lim_{x\to\infty}f(x) = \lim_nf(x_n) = f(a)$, because limit of constant sequence is equal to that constant. Since $a$, was arbitrary, we get $f(a) = A$, for all $a\in(0,\infty)$. – 2017-02-08
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Further thoughts and a little problem:

Interesting. This is similar to (but not the same as) a periodic function.

In a periodic function f(x + k) = f(x) for all x. For example in f(x) = sin(x), the minimum k = 2$\pi$ so the period is 2$\pi$.

Here we have f(cx) = f(x) with c = 2. The "periods" increase geometrically, with f(x) = f(2x) = f(4x) = f(8x) $\ldots$

One example of such a function would be $$f(x) = sin(\frac{2\pi}{log(2)} log ( x))$$
Then $$f(2x) = sin(\frac{2\pi}{log(2)} log( 2x))$$ $$ = sin(\frac{2\pi}{log(2)} (log( 2) +log(x)))$$ $$ = sin( 2\pi +\frac{2\pi}{log(2)}log(x))$$ $$ = sin( \frac{2\pi}{log(2)}log(x))$$ $$ = f(x)$$

This is more detail than strictly required by the question but it is an interesting pattern and worth exploring.

In the above function the value of f(x) keeps varying between 1 and -1 even as the "periods" get longer exponentially. If as given in the question $\lim_{x\rightarrow \infty} f(x) = A$ then the above function would not work.

But now what would be wrong with this function, which approaches zero as x tends to infinity?

$$ f(x) = e^{-x} sin( \frac{2\pi}{log(2)}log(x))$$

The above function is defined on $(0, \infty)$ as required [If we were required to use a closed interval that would be a problem for log but as given above the interval is open] and it seems to satisfy the requirements given. But it is not constant.

Thoughts anyone?

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    $$f(2x) = e^{-2x}\sin\left(\frac{2\pi}{\ln 2}\ln (2x)\right) = e^{-2x}\sin\left(\frac{2\pi}{\ln 2}\ln (x)\right) \neq f(x)$$ – 2017-02-08